+0

# 2016 NS 28

0
356
12
+935

Can anyone solve this?

Mar 13, 2019

#1
+12
-1

how did you upload a picture it wont worrk for me cant figure it out msg me

Mar 13, 2019
#2
+1

sumfor(n, 1, 2016, (n^2) = 2733212496 mod 17 = 11

Mar 13, 2019
#3
+110068
+1

What calc did you put that into?

Mar 13, 2019
#4
+1

Hello Melody: Mathematica 11 Home Edition. I actually expressed it like this: ∑[n^2, n, 1, 2016] mod 17 = 11.

But, I also re-checked it using C++ programming language.

Guest Mar 13, 2019
edited by Guest  Mar 13, 2019
#6
+110068
+1

Thanks

Melody  Mar 14, 2019
#5
+6194
+1

$$\left .\sum \limits_{k=1}^n ~k^2 = \dfrac 1 6 n (1+n)(1+2n) \right |_{n=2016} = \\ \dfrac 1 6 (2016)(2017)(4033)$$

$$\dfrac 1 6 (2016)(2017)(4033) \pmod{17} = \\ (336 \pmod {17})(2017 \pmod{17})(4033 \pmod{17}) \pmod{17} = \\ 13 \cdot 11 \cdot 4 \pmod{17} = \\ 572 \pmod{17} = 11$$

.
Mar 13, 2019
edited by Rom  Mar 13, 2019
edited by Rom  Mar 14, 2019
#7
+110068
+1

Thanks Rom.

Melody  Mar 14, 2019
#8
+935
0

Wait, how did you get the first summation? Is there a formula or something?

dgfgrafgdfge111  Mar 14, 2019
#9
+6194
+1

Yeah.  That's a pretty standard result.

Rom  Mar 14, 2019
#11
+935
-1

So what's the formula?

dgfgrafgdfge111  Mar 14, 2019
#12
+935
0

Oh, I get it now. I see. Thanks a lot for your help (and time)!!!

dgfgrafgdfge111  Mar 14, 2019
edited by dgfgrafgdfge111  Mar 14, 2019
#10
+25451
+2

2016 NS 28

$$\begin{array}{|ll|} \hline 0^2+1^2+2^2+\ldots +7^2+8^2 &+9^2 +10^2+\ldots +15^2+16^2 \\ +17^2+18^2+19^2+\ldots +24^2+25^2 &+26^2+27^2\ldots +32^2+33^2 \\ \dots \\ +2006^2+2007^2+2008^2+\ldots +2013^2+2014^2 &+2015^2+2016^2 \pmod{17}\\ =\\ \color{red}\underbrace{0^2+1^2+2^2+\ldots+7^2+8^2}_{=0\pmod{17}}&\color{red}\underbrace{+9^2 +10^2+11^2+\ldots +15^2+16^2}_{=0\pmod{17}} \\ \color{red}\underbrace{+0^2+1^2+2^2+\ldots+7^2+8^2}_{=0\pmod{17}}&\color{red}\underbrace{+9^2 +10^2+11^2+\ldots +15^2+16^2}_{=0\pmod{17}} \\ \color{red}\dots \\ \color{red}\underbrace{+0^2+1^2+2^2+\ldots+7^2+8^2}_{=0\pmod{17}}&\color{red}+9^2+10^2 \pmod{17} \\\\ =9^2+10^2 \pmod{17}\\ =11 \pmod{17}\\ \hline \end{array}$$

Mar 14, 2019